ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:06Z
∑_{k=1}^{n} k = n(n+1)/2
4β² + 5β + 0 = 0
∫₀^∞ e^(-β²) dβ = √π / 2
125K
∑_{k=1}^{n} k = n(n+1)/2
91K
d/dφ [φ^6] = 6φ^5
6n² + 3n + 0 = 0
10β² + 6β + 6 = 0
F = ma
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
364K